Develops the sequences whose consecutive terms have a constant ratio, presenting them as the multiplicative counterpart of arithmetic sequences. Covers the exponential term formula, the sum formula and its derivation, and the distinction between growth and decay.
Subject: Precalculus · 65 slides · symbolic lesson
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Title
Precalculus · Chapter 11 — Sequences, Probability and Counting Theory
§11.3 Geometric Sequences, pp. 1320-1330
Objectives
Five things, each one you can check yourself on paper.
OpenStax, Precalculus, §11.3 Geometric Sequences §11.3, pp. 1320-1330 — the pages these objectives are drawn from
Warm-up
The previous section added a fixed amount at each step.
Discussion prompt
In the list 3, 6, 12, 24, what happens at each step?
Hint: The differences are not constant.
Answer:
The differences are 3, 6 and 12 — not constant, so this is not arithmetic.
But each term is exactly twice the previous one. The ratio is constant even though the difference is not.
So there is a second family: constant multiplication rather than constant addition. Everything from the last section translates, with addition replaced by multiplication throughout.
Concept
A geometric sequence multiplies by the same amount at each step. Because the change is proportional, the explicit formula involves a power rather than a multiple.
common ratio — the constant factor by which each term of a geometric sequence is multiplied to get the next
\[ a_n=a_1r^{n-1} \]
The exponent is one less than the index, for exactly the reason the arithmetic formula's coefficient was: the first term is reached without any steps. Here the off-by-one hides in an exponent, which makes it easier to lose.
Figure (svg): A sequence of terms with the constant ratio between consecutive pairs marked by arrows
OpenStax, Precalculus, §11.3 Geometric Sequences §11.3, pp. 1320-1323
Section
Section 1
Concept
Testing whether a sequence is geometric means dividing each term by the one before and confirming the quotients all agree.
Dividing in the right order matters as much as subtracting in the right order did. Later divided by earlier gives the factor applied going forwards, and reversing it gives the reciprocal, which describes the sequence backwards.
Figure (svg): A sequence of terms with the constant ratio between consecutive pairs marked by arrows
OpenStax, Precalculus, §11.3 Geometric Sequences §11.3, pp. 1320-1324
Picture it
Each arrow multiplies by the same amount.
Figure (svg): A sequence of terms with the constant ratio between consecutive pairs marked by arrows
The bottom line names the consequence. Repeated multiplication produces a power, so the explicit formula is exponential rather than linear.
Worked example
Compute all the ratios.
\[ \text{Is } 81,\;27,\;9,\;3 \text{ geometric?} \]
First ratio
Why: Second over first.
\[ \frac{1}{3} \]
Second ratio
Why: Third over second.
\[ \frac{1}{3} \]
Third ratio
Why: Fourth over third.
\[ \frac{1}{3} \]
Compare
Why: All equal.
Figure (svg): A sequence of terms with the constant ratio between consecutive pairs marked by arrows
\[ r=\tfrac{1}{3} \]
Verify: check the direction
Why: A ratio below one means the terms shrink, which they do. Dividing in the other order would have given 3 and described the sequence read backwards, which is a different sequence.
OpenStax, Precalculus, §11.3 Geometric Sequences §11.3, pp. 1321-1323
Prediction
A sequence has terms 4, 12, 36.
Predict first
What comes next?
Correct: One hundred and eight.
Why: Each term is three times the previous one, so the next is 36 times 3. The differences grow, which is what rules out an arithmetic reading.
Worked example
Alternating signs.
\[ \text{Find the common ratio of } 2,\;-6,\;18,\;-54. \]
First ratio
Why: Second over first.
\[ -3 \]
Second ratio
Why: Third over second.
\[ -3 \]
Confirm
Why: Constant.
Note the effect
Why: Signs alternate.
Figure (svg): The solution to Worked example a negative ratio shown as a ladder of expressions, one row per legal move
\[ r=-3 \]
Verify: check what a negative ratio does
Why: Multiplying by a negative number flips the sign every step, so the terms alternate. The sizes still grow by a factor of three, so the sequence both alternates and grows — two effects from one negative ratio.
OpenStax, Precalculus, §11.3 Geometric Sequences §11.3, pp. 1323-1324
Trap
\[ 3,\;6,\;12: \text{ differences }3,\;6 \;\Longrightarrow\; \text{no pattern} \]
Conclude there is no pattern from the differences alone
Why: Only the arithmetic test is applied.
The constant ratio is missed and a perfectly regular sequence is called irregular.
Test both: differences for arithmetic and ratios for geometric.
A sequence failing one test may pass the other, and these two families cover most of what this chapter uses.
Both tests take seconds, so running both before concluding anything is the reliable habit.
Faded example
From two consecutive terms.
Fill in the blanks
r=\frac3shrink=\frac______}, \text______
Why: Later divided by earlier gives the factor applied going forwards. A ratio below one in size means each term is smaller than the last.
Sorting
Test differences and ratios.
Sort into buckets
Sort each sequence.
Step zero
You are given a list and asked to identify the family.
Discussion prompt
What do you compute?
Hint: Two tests, not one.
Answer:
The differences and the ratios. A constant difference means arithmetic and a constant ratio means geometric.
Running only one test risks concluding there is no pattern when the other would have found it — a sequence that doubles has no constant difference at all.
Both take seconds, and between them they cover the two families this chapter develops. Neither being constant means the sequence belongs to neither family, which is itself a useful conclusion.
Section
Section 2
Concept
Every idea from the previous section has a geometric counterpart obtained by replacing addition with multiplication and multiplication with exponentiation.
The translation is exact enough that a formula from either section can be recovered from the other. That makes the second family much less to learn than it appears, provided the correspondence is noticed early.
| arithmetic | geometric |
|---|---|
| common difference | common ratio |
| add at each step | multiply at each step |
| linear explicit formula | exponential explicit formula |
| multiply the difference by the step count | raise the ratio to the step count |
| graph is dots on a line | graph is dots on an exponential curve |
Figure (svg): A contrast between arithmetic and geometric sequences, showing the term-by-term translation between them
OpenStax, Precalculus, §11.3 Geometric Sequences §11.3, pp. 1324-1326
Picture it
Each row translates directly.
Figure (svg): A contrast between arithmetic and geometric sequences, showing the term-by-term translation between them
The caption flags where the translation makes an error easier. An off-by-one in an exponent is much less visible than one in a coefficient.
Worked example
Multiplication becomes exponentiation.
\[ \text{Translate } a_n=a_1+(n-1)d \text{ into its geometric form.} \]
Translate the operation
Why: Adding becomes multiplying.
Translate the repetition
Why: Repeated addition becomes repeated multiplication.
Keep the step count
Why: Still one fewer than the index.
\[ \text{exponent } n - 1 \]
Write the result
Why: The geometric term formula.
\[ a _{1} r ^{n - 1} \]
Figure (svg): A contrast between arithmetic and geometric sequences, showing the term-by-term translation between them
\[ a_n=a_1r^{n-1} \]
Verify: check the first term
Why: At index one the exponent is zero, and any nonzero number to the power zero is one — so the formula gives the first term exactly. That mirrors the arithmetic case, where the coefficient was zero at the first index.
OpenStax, Precalculus, §11.3 Geometric Sequences §11.3, pp. 1324-1325
Matching
Every idea translates.
Match the pairs
Why: Each geometric idea is its arithmetic counterpart with the operation raised one level: addition becomes multiplication and multiplication becomes exponentiation. That correspondence makes the second family much less to learn.
Worked example
Constant amount or constant percentage.
\[ \text{A population grows by } 200 \text{ a year against one growing by } 5\% \text{ a year. Which is which?} \]
Consider the fixed amount
Why: The same number added.
Classify it
Why: Arithmetic.
Consider the percentage
Why: The same factor applied.
Classify it
Why: Geometric.
Figure (svg): The solution to Worked example which family models this shown as a ladder of expressions, one row per legal move
\[ \text{amount: arithmetic; percent: geometric} \]
Verify: check what a percentage means
Why: Growing by five per cent means multiplying by 1.05, which is a constant factor rather than a constant addition. The amount added grows each year even though the percentage does not — which is exactly the distinction between the two families.
OpenStax, Precalculus, §11.3 Geometric Sequences §11.3, pp. 1325-1326
Error analysis
A student writes a geometric term formula.
Annotate
On: \( a_n=a_1\cdot r\cdot(n-1) \)
This is the translation applied incompletely: the step count carried over but the operation did not. Checking a second term against the sequence catches it immediately.
Prediction
A quantity grows by three per cent each year.
Predict first
Which family models it?
Correct: Geometric, since a percentage is a constant factor.
Why: Growing by three per cent means multiplying by 1.03 each year, which is a constant ratio. The amount added grows each year, so the differences are not constant.
Faded example
From arithmetic to geometric.
Fill in the blanks
a_1+(n-1)d \;\to\; a_1\cdot r^1}}, \text1n-___
Why: The step count survives the translation unchanged; only the operation changes. Repeated addition of a difference becomes repeated multiplication by a ratio, which is a power.
Explain it
The two families are closely parallel.
Discussion prompt
Explain the correspondence to a classmate.
Hint: Which operation replaces which?
Answer:
Everywhere the arithmetic family adds, the geometric family multiplies. The common difference becomes a common ratio.
And everywhere the arithmetic family multiplies — the difference by the step count — the geometric family exponentiates, raising the ratio to that count.
So the second family is the first with every operation raised one level. A good explanation notes that this halves what has to be learned, provided the correspondence is noticed rather than treating the sections as independent.
Section
Section 3
Concept
Reaching the nth term takes one fewer multiplication than the index, for exactly the reason the arithmetic formula took one fewer addition.
The exponent being zero at the first index is the check, and it works because any nonzero base to the power zero is one. That mirrors the arithmetic case, where the coefficient of the difference was zero there.
Figure (svg): A contrast between arithmetic and geometric sequences, showing the term-by-term translation between them
OpenStax, Precalculus, §11.3 Geometric Sequences §11.3, pp. 1326-1328
Picture it
As a coefficient on the left and an exponent on the right.
Figure (svg): A contrast between arithmetic and geometric sequences, showing the term-by-term translation between them
An exponent that is one too large multiplies the answer by the whole ratio, which is a much bigger error than adding one extra difference — so the check matters more here.
Worked example
One exponentiation.
\[ \text{Find the tenth term of } 3,\;6,\;12,\;\ldots \]
Identify the first term
Why: From the list.
\[ 3 \]
Identify the common ratio
Why: Later over earlier.
\[ 2 \]
Count the steps
Why: One fewer than ten.
\[ 9 \]
Compute
Why: Three times two to the ninth.
\[ 3 \times 512 \]
Figure (svg): A contrast between arithmetic and geometric sequences, showing the term-by-term translation between them
\[ a_{10}=1536 \]
Verify: check the first term
Why: At index one the exponent is zero, giving three times one, which is 3 — the correct first term. An exponent of ten instead of nine would have doubled the answer, which is why the check matters more here than in the arithmetic case.
OpenStax, Precalculus, §11.3 Geometric Sequences §11.3, pp. 1326-1327
Prediction
Finding the fifteenth term.
Predict first
To what power is the ratio raised?
Correct: Fourteen.
Why: The first term takes no multiplications, so reaching the fifteenth takes fourteen. Using fifteen would multiply the answer by an extra factor of the ratio.
Worked example
The exponent difference does the work.
\[ \text{A geometric sequence has } a_2=6 \text{ and } a_5=48. \text{ Find the ratio.} \]
Count the steps between them
Why: Five minus two.
\[ 3\text{ steps} \]
Write the relationship
Why: Three multiplications.
\[ 48 = 6 r ^{3} \]
Isolate the power
Why: Divide.
\[ r ^{3} = 8 \]
Take the cube root
Why: The ratio.
\[ r = 2 \]
Figure (svg): The solution to Worked example find the ratio from two terms shown as a ladder of expressions, one row per legal move
\[ r=2 \]
Verify: check by generating
Why: From 6 the terms run 12, 24, 48 — reaching 48 at the fifth position as required. Counting the steps between the two given positions, rather than using either index directly, is what makes this work.
OpenStax, Precalculus, §11.3 Geometric Sequences §11.3, pp. 1327-1328
Trap
\[ a_{10}=3\cdot 2^{10}=3072 \]
Raise the ratio to the index
Why: The step count is taken to equal the index.
The answer is exactly one ratio too large — double, here.
The exponent is one less than the index, because the first term takes no multiplications.
At the first index the exponent must be zero, giving the first term back.
The error is multiplicative here, so it doubles or triples the answer rather than shifting it slightly — which makes the check worth doing every time.
Faded example
First term 3, ratio 2, tenth term.
Fill in the blanks
a_1=3\cdot 2^512}}=3\cdot___=1536
Why: Nine multiplications by two give 512, and the first term multiplies that. The minus one converts a term count into a step count exactly as it did in the arithmetic formula.
Sorting
Substitute the first index.
Sort into buckets
Sort each formula for a sequence starting at 3 with ratio 2.
Explain it to yourself
The same off-by-one appeared in the previous section.
Discussion prompt
Explain why it is worse in an exponent.
Hint: What does the error do to the answer?
Answer:
In the arithmetic formula an extra step adds one common difference — a small absolute error that is often visible as a slightly wrong answer.
In the geometric formula an extra step multiplies by the whole ratio, so the answer can be double or triple what it should be.
And an exponent is less visible than a coefficient when scanning a formula. A good explanation concludes that the substitution check should be automatic here, since the error is both larger and harder to spot.
Section
Section 4
Concept
Multiplying the sum by the common ratio shifts every term along by one position, so subtracting cancels everything except the two at the ends.
The exclusion of a ratio of one is not a technicality: a ratio of one gives a constant sequence, whose sum is simply the count times the term. The formula's denominator would be zero, and the special case is handled separately.
Figure (svg): A card showing the derivation of the geometric sum formula by subtracting the sum from a multiple of itself
OpenStax, Precalculus, §11.3 Geometric Sequences §11.3, pp. 1328-1330
Picture it
One multiplication and one subtraction.
Figure (svg): A card showing the derivation of the geometric sum formula by subtracting the sum from a multiple of itself
The two lines are offset by one position, which is why almost everything cancels. Seeing the derivation once makes the formula reconstructible rather than memorised.
Worked example
Substitute into the formula.
\[ \text{Sum the first } 8 \text{ terms of } 3,\;6,\;12,\;\ldots \]
Identify the pieces
Why: First term, ratio, count.
\[ 3, 2, 8 \]
Compute the ratio's power
Why: Two to the eighth.
\[ 256 \]
Form the numerator
Why: First term times one minus that.
\[ 3(1 - 256) \]
Divide by one minus the ratio
Why: And simplify.
\[ -765 / - 1 \]
Figure (svg): A card showing the derivation of the geometric sum formula by subtracting the sum from a multiple of itself
\[ S_8=765 \]
Verify: check by adding a few
Why: The terms are 3, 6, 12, 24, 48, 96, 192, 384, and adding them gives 765 — matching. The formula reached it without adding eight numbers, which is the point of having it.
OpenStax, Precalculus, §11.3 Geometric Sequences §11.3, pp. 1328-1329
Prediction
Summing the first eight terms.
Predict first
To what power is the ratio raised?
Correct: Eight, the term count.
Why: The sum formula uses the count itself, unlike the term formula which uses one less. That difference is a genuine trap once the term formula's minus one has become automatic.
Worked example
Multiply, subtract, solve.
\[ \text{Derive the sum formula for a geometric sequence.} \]
Write the sum
Why: All the terms.
Multiply by the ratio
Why: Every term shifts up one power.
Subtract
Why: The shared terms cancel.
\[ S - r S = a - a r ^{n} \]
Factor and divide
Why: Solve for the sum.
\[ S = a(1 - r ^{n}) / (1 - r) \]
Figure (svg): The solution to Worked example follow the derivation shown as a ladder of expressions, one row per legal move
\[ S_n=\frac{a_1(1-r^n)}{1-r} \]
Verify: check why the ratio cannot be one
Why: With a ratio of one the denominator is zero and the formula fails. But that case is a constant sequence, whose sum is just the count times the term — so the exclusion removes a case that needs no formula at all.
OpenStax, Precalculus, §11.3 Geometric Sequences §11.3, pp. 1329-1330
Error analysis
A student sums eight terms.
Annotate
On: \( S_8=\frac{3(1-2^7)}{1-2} \)
The two formulas use the exponent differently, which is a genuine trap after the term formula's minus one has been drilled in. Adding the first few terms by hand and comparing is the quickest check.
Faded example
First term 3, ratio 2, eight terms.
Fill in the blanks
S=\frac8}})}765=\frac______=___
Why: The exponent is the term count and both the numerator and denominator come out negative, so the quotient is positive. Checking against a short hand sum confirms it.
Sorting
The two formulas differ here.
Sort into buckets
Sort each formula.
Explain it
The sum formula comes from one clever step.
Discussion prompt
Explain it to a classmate.
Hint: What does multiplying by the ratio do?
Answer:
Multiplying the sum by the ratio shifts every term along by one position, so the new line and the old one share almost all their terms.
Subtracting cancels every shared term, leaving only the first term from one line and one extra term from the other.
Solving the result for the sum gives the formula. A good explanation notes why the ratio cannot be one: the subtraction would give zero equals zero, and that case is a constant sequence needing no formula at all.
Section
Section 5
Concept
A ratio larger than one in size makes the terms grow without bound; a ratio smaller than one makes them shrink towards zero.
The decay case is the one that matters most in the next section: terms shrinking towards zero are exactly what allows infinitely many of them to have a finite total. Growth makes an infinite sum impossible.
| ratio | behaviour |
|---|---|
| greater than one | grows without bound |
| between zero and one | decays towards zero |
| exactly one | constant |
| negative, size above one | alternates and grows |
| negative, size below one | alternates and decays |
Figure (svg): Two geometric sequences plotted, one growing with a ratio above one and one decaying with a ratio below one
OpenStax, Precalculus, §11.3 Geometric Sequences §11.3, pp. 1322-1330
Picture it
Two sequences, distinguished by their ratios.
Figure (svg): Two geometric sequences plotted, one growing with a ratio above one and one decaying with a ratio below one
The bottom line points forward. Shrinking terms are what make the infinite sums of the next section possible, and growing terms rule them out.
Worked example
A percentage means a ratio.
\[ \text{An investment of } \$1000 \text{ grows } 6\% \text{ a year. Find its value after } 10 \text{ years.} \]
Convert the percentage
Why: Add to one.
\[ r = 1.06 \]
Set the first term
Why: The initial amount.
\[ 1000 \]
Count the multiplications
Why: Ten years means ten.
\[ \text{exponent } 10 \]
Compute
Why: One thousand times the power.
\[ \text{about } 1790 \]
Figure (svg): Two geometric sequences plotted, one growing with a ratio above one and one decaying with a ratio below one
\[ \approx\$1790 \]
Verify: check the indexing
Why: Ten years of growth means ten multiplications, so the exponent is ten rather than nine — the initial amount is at year zero rather than being term one. Stating that correspondence explicitly is what keeps the exponent right.
OpenStax, Precalculus, §11.3 Geometric Sequences §11.3, pp. 1325-1328
Prediction
A quantity falls by 40 per cent each period.
Predict first
What is the common ratio?
Correct: Zero point six.
Why: Losing forty per cent leaves sixty per cent, and the ratio is what remains rather than what is lost. Using 0.4 would model keeping only forty per cent, which is a much faster decay.
Worked example
A ratio below one.
\[ \text{A drug's concentration falls } 25\% \text{ each hour from } 80 \text{ mg. Find it after } 5 \text{ hours.} \]
Convert the percentage
Why: Falling by a quarter leaves three quarters.
\[ r = 0.75 \]
Set the first term
Why: The initial amount.
\[ 80 \]
Count the multiplications
Why: Five hours.
\[ \text{exponent } 5 \]
Compute
Why: Eighty times the power.
\[ \text{about } 19 \]
Figure (svg): The solution to Worked example model decay shown as a ladder of expressions, one row per legal move
\[ \approx 19\text{ mg} \]
Verify: check the ratio's direction
Why: Falling by 25 per cent means keeping 75 per cent, so the ratio is 0.75 rather than 0.25. Using 0.25 would model losing three quarters each hour instead of a quarter, giving a far smaller answer — which is the standard slip with decay percentages.
OpenStax, Precalculus, §11.3 Geometric Sequences §11.3, pp. 1328-1330
Trap
\[ \text{falls }25\% \;\Longrightarrow\; r=0.25 \]
Take the percentage change as the ratio
Why: The stated loss is used directly as the multiplier.
That models keeping a quarter rather than losing a quarter.
The ratio is what remains, not what is lost. Losing a quarter leaves three quarters, so the ratio is 0.75.
For growth the same logic gives one plus the rate, since the original plus the increase is kept.
Ask what fraction survives each period, which is the ratio, rather than what fraction changes.
Sorting
Compare the ratio's size with one.
Sort into buckets
Sort each ratio.
Faded example
Six per cent annual growth.
Fill in the blanks
r=1+0.06=1.06
Why: The ratio is what remains after the period, which for growth is the original plus the increase. For decay it is the original minus the loss, giving a ratio below one.
Explain it
The next section sums infinitely many terms.
Discussion prompt
Explain to a classmate why the ratio's size will matter there.
Hint: What must the terms do?
Answer:
For infinitely many terms to have a finite total, the terms must shrink towards zero — otherwise the sum grows without bound.
A ratio below one in size makes each term a fraction of the last, so the terms shrink geometrically and the additions become negligible.
A ratio at or above one makes the terms stay the same or grow, and no infinite sum is possible. A good explanation notes that the ratio's size is therefore the whole criterion for whether §11.4's infinite sums exist.
Comparison
Fill the blanks from memory. Every row translates.
Comparison matrix
| arithmetic | geometric | |
|---|---|---|
| what is constant | the difference | the ratio |
| each step | adds | multiplies |
| explicit formula | linear | exponential |
| the off-by-one appears as | a coefficient | an exponent |
The last row is where the translation makes an error easier. An exponent one too large multiplies the answer by the ratio, where a coefficient one too large adds only one difference.
Pattern
Five steps, and the last catches the section's costliest error.
Step 4 is the one place the two formulas genuinely differ: the term formula's exponent is one less than the index and the sum formula's is the count itself.
OpenStax Algebra and Trigonometry 2e, §13.3 Geometric Sequences §13.3
Check
The term formula.
Check your understanding
In the term formula, what is the ratio's exponent?
Answer: A
Why: The first term is reached without any multiplications, so the exponent is zero there and one less than the index generally. Using the index itself multiplies every answer by an extra factor of the ratio.
Check
Modelling decay.
Check your understanding
A quantity falls by 20 per cent each period. What is the common ratio?
Answer: A
Why: The ratio is what remains, not what is lost. Losing a fifth leaves four fifths, so each term is 0.8 times the last.
Check
Identifying the family.
Check your understanding
What makes a sequence geometric?
Answer: A
Why: A constant ratio is the defining property, and it makes the explicit formula exponential. A constant difference defines the arithmetic family instead.
Real world
Half-life is a geometric sequence with a ratio of one half.
Discussion prompt
A radioactive sample halves every 5730 years. Why is that geometric rather than arithmetic?
Hint: What stays the same at each step?
Answer:
The fraction remaining is constant, not the amount lost. Each period leaves half of whatever was there, so the ratio is one half.
The amount lost shrinks each period — a large sample loses far more than a small one — which is exactly why the differences are not constant.
Radiocarbon dating inverts the formula: measuring what fraction remains and solving for the number of periods gives the age. That solving step is a logarithm, which is why chapter 4's material and this section's meet in the same computation.
Commit first
State your confidence along with your answer.
Predict first
Why is a percentage change modelled geometrically rather than arithmetically?
Correct: A percentage is a constant factor, not a constant amount.
Why: Growing by five per cent means multiplying by 1.05 each period, so the ratio is constant while the amount added grows. A constant amount added each period would be arithmetic instead.
Explain it
The test of understanding is being able to say why, not just what.
Discussion prompt
A classmate says a sequence has no pattern because its differences are not constant. What do you suggest?
Hint: There is a second test.
Answer:
Compute the ratios as well. A sequence with no constant difference may well have a constant ratio, and then it is geometric.
Doubling terms have differences of 3, 6, 12 and so on — clearly not constant — but a ratio of exactly two throughout.
So both tests are needed before concluding there is no pattern. A good explanation adds that the two families cover most of what this chapter uses, so running both takes seconds and settles the question.
Exit ticket
One honest answer, so the next lesson can start in the right place.
Predict first
Which idea from this lesson would you most want to see again?
Correct: Any of these is a legitimate answer; the useful one is the honest one.
Why: There is no correct choice here. The third is where the costliest error lives, since an exponent one too large multiplies the answer. The second is what makes the whole section half as much to learn.
Connect it up
One page, drawn from memory, is worth more than rereading the section.
Draw it
Write a two-column table translating every arithmetic idea into its geometric counterpart. Beside it, write both geometric formulas and mark clearly which exponent is the index and which is one less. Underneath, derive the sum formula by multiplying and subtracting, and model one growth and one decay application.
If your two formulas have their exponents distinguished explicitly, and your decay model uses what remains rather than what is lost, the section's two costly errors are both accounted for.
Recap
Five things, and the fourth is the one to check every time.
| if you remember one thing | it should be this |
|---|---|
| about the family | constant ratio, so an exponential formula |
| about the term formula | exponent one less than the index, zero at the first |
| about the sum formula | exponent is the count, not one less |
| about percentages | the ratio is what remains, not what changes |
Section 11.4 introduces summation notation and then asks what happens when a geometric sequence is summed forever — which is finite precisely when the terms decay.
OpenStax, Precalculus, §11.3 Geometric Sequences §11.3, pp. 1320-1330 — everything on these slides traces back here
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